What is the solution for cos(ax)/a when a approaches 0?

[e_stoimenov]

Hi all ,
I'm an electronics engineer and I'm trying to solve this two equations:

1.
sin(ax)/a, when a=0
I do it this way:
Multiply by x and make a limit transition
lim sin(ax)x/ax, as lim sin(ax)/ax = 1
ax->0 ax->0
I infer:
lim sin(ax)x/ax = x
ax->0

2.
Unfortunatelly I'm nowhere with
cos(ax)/a, when a=0

Any help :)
P.S.
Sorry for the bad english.

 

[CompuChip]

Are you allowed to use L'Hopital's rule?
Since your first limit is of the from
$$\lim_{a \to 0} f(a) / g(a)$$ where $$\lim_{a \to 0} f(a) = \lim_{a \to 0} g(a) = 0$$
it would apply in this case.

The second one is easier, because it is of the from
$$\lim_{a \to 0} f(a) / g(a)$$ where $$\lim_{a \to 0} f(a)$$ exists and is finite, i.e. it is similar to
$$\lim_{a \to 0} f(0) / g(a)$$

 

[HallsofIvy]

Hi all ,
I'm an electronics engineer and I'm trying to solve this two equations:

1.
sin(ax)/a, when a=0
I do it this way:
Multiply by x and make a limit transition
lim sin(ax)x/ax, as lim sin(ax)/ax = 1
ax->0 ax->0
I infer:
lim sin(ax)x/ax = x
ax->0

That is, as you say, a limit. It is NOT the value of sin(ax)/a "when a= 0". That has no value.

2.
Unfortunatelly I'm nowhere with
cos(ax)/a, when a=0

Any help :)
P.S.
Sorry for the bad english.

That is because the limit does not exist. The denominator is going to 0 as the numerator goes to 1. As a gets closer and closer to 0, the absolute value gets larger and larger.

 

[e_stoimenov]

Hi, thanks for your replyes!

Dear CompuChip,
I really haven't investigate if the N1 limit
excits at all, so I can apply the L'hopital rule.

"The second one is easier", ok, that encaurages me!

Dear HallsofIvy,
I have to admite that I'm making a limit transition,
so I'm searching for the limit, but not for the exact value.

Thanks once again!

 

[zetafunction]

the cosine is BOUNDED it can not be greater than 1 ,however 1/a is UNBOUNDED so the limit is infinite

if you expand the cosine(ax) into a taylor series in (ax) there is no linear part on 'x' only a constant term 1 and a quadratic term so there is no cancellation of the linear term and the limit is infinite

 

[CompuChip]

I don't know if this is legal, but I just made it up:
You can multiply by sin(ax)/sin(ax) (the limit point a = 0 is not in the domain we are considering anyway) and write it as
$$\frac{\cos(ax)}{\sin(ax)} \cdot \frac{\sin(ax)}{a}$$
Now if you take the limit, the second factor is a standard limit (which you have just proven) going to one, and the first factor is 1/tan(a x). As the reciprocal of a function continuous at a x = 0 this limit does not exist, and therefore neither does the original limit.

(Of course, this sort of hinges on the fact that you can calculate $$\lim_{x \to 0} f(x)$$ as $$\lim_{x \to 0} f(x) g(x)$$ when [itex]\lim_{x \to 0} g(x) = 1[/tex]).[/itex]

 

[e_stoimenov]

Dear CampuChip,
Your solution is very interesting.
According to my first post the
lim sin(ax)/x = x, when ax->0,
but not to 1.
So if we put this into your solution:
lim cos(ax)/x = lim [cos(ax).x/sin(ax)], when ax->0.
10x,
Eltimir

 

[CompuChip]

Yes, sorry, my bad.
You showed that
$$\lim_{a \to 0} \frac{\sin ax}{a} = x$$
I claimed that
$$\lim_{a \to 0} \frac{\sin a}{a} = 1$$
(which is a standard limit) and mistakenly said that that is what you had proven.